I have two questions regarding parametric equations that I am struggling with.
Question 1:
a) Give a parametric equation for the line passing through $(-1,-2,3)$ and $(1,5,-2)$.
b) Give the symmetric equation for the line in part a).
My work:
I have found a parametric equation for part a), namely:
$x = -1+2t \\ y = 5+7t \\ z = -2-5t$
I am not sure what the symmetric equation is, or even what a symmetric equation is for that matter.
Question 2:
Consider the plane $Q$, defined by the equation $2x-2y+z=15$ and the point $P$ with coordinates $(11,-7,6)$.
a) Find a parametric equation for the line through $P$ normal to $Q$.
b) Where does this normal line through $P$ intersect $Q$?
c) What is the distance from $P$ to $Q$?
My Work I was able to figure out part c) easily, finding that the distance is $7$. That being said, I'm not sure how to solve parts a) and b).
Question 1:
The symmetric equation of a line is in the form of
$$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$$
In your case, $\frac{x+1}{2}=t$, what else are equal to $t$? Combine them as above, you will get the symmetric equation.
Question 2:
(a) The parametric equation for a line is $\vec{x}=\vec{P}+\vec{d}t$, where $\vec{P}$ is a point on the line, $\vec{d}$ is a direction vector for the line. In this case, the direction vector for the line is the normal vector for the plane, so $(2,-2,1)$. Can you figure out the rest then?
(b) Once you find the line, it should be easy to find the intersection.